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AC peak and RMS values; reactance and impedance: NEET MCQs with answers

59 questions on this topic of Electromagnetic Induction and Alternating Currents (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A capacitor is connected to an AC source of voltage V and frequency f. If the frequency of the source is reduced to f/3 while maintaining the rms voltage constant, the displacement current through the capacitor becomes:

  1. Three times its initial value
  2. Nine times its initial value
  3. One-third of its initial value
  4. One-ninth of its initial value
Show answer and explanation

Answer: (C) One-third of its initial value

The conduction current equals the displacement current in the capacitor circuit: I = V / X_C = V(2πfC). Since I is directly proportional to frequency f, reducing the frequency to one-third causes the current to decrease to one-third of its initial value.

medium

2. The inductive reactance of a 20 mH inductor at 50 Hz is:

  1. 0.159 Ω
  2. 6280 Ω
  3. 1 Ω
  4. 6.28 Ω
Show answer and explanation

Answer: (D) 6.28 Ω

X_L = 2πfL = 2π × 50 × 20 × 10⁻³ = 6.28 Ω.

hard

3. A symmetrical periodic current wave has a constant value of +I₀ for the first third of the time period (0 to T/3), zero for the next third (T/3 to 2T/3), and -I₀ for the final third (2T/3 to T). The rms value of this current over one complete cycle is:

  1. I₀ / √3
  2. I₀ √(1/3)
  3. I₀ / 2
  4. I₀ √(2/3)
Show answer and explanation

Answer: (D) I₀ √(2/3)

The mean square value is defined as <i²> = (1/T) ∫₀ᵀ i² dt. Breaking into intervals: ∫₀^(T/3) I₀² dt + ∫_(T/3)^(2T/3) 0 dt + ∫_(2T/3)^T (-I₀)² dt = I₀²(T/3) + 0 + I₀²(T/3) = 2I₀²T/3. Thus, <i²> = (1/T) × (2I₀²T/3) = (2/3)I₀². Taking the square root gives I_rms = I₀√(2/3).

medium

4. An alternating current is given by the equation I = 6 sin(100πt) + 8 cos(100πt) A. The root mean square (rms) value of this current is:

  1. 5 A
  2. 7 A
  3. 5√2 A
  4. 10 A
Show answer and explanation

Answer: (C) 5√2 A

The resultant peak current is I₀ = √(6² + 8²) = √100 = 10 A. The rms value of the sinusoidal current is given by I_rms = I₀ / √2 = 10 / √2 = 5√2 A.

easy

5. A coil of resistance 30 Ω and inductance 0.4 H is connected to an AC source of 200 V, 50/π Hz. The impedance of the coil is:

  1. 30 Ω
  2. 40 Ω
  3. 70 Ω
  4. 50 Ω
Show answer and explanation

Answer: (D) 50 Ω

The inductive reactance is X_L = 2πfL = 2π × (50/π) × 0.4 = 40 Ω. The impedance of the coil is Z = √(R² + X_L²) = √(30² + 40²) = 50 Ω.

hard

6. A direct current of 4 A flows through a resistor. Simultaneously, an alternating current given by I = 3√2 sin(ωt) A is passed through the same resistor. The effective (rms) value of the combined current is:

  1. 7 A
  2. 5 A
  3. (4 + 3√2) A
  4. √7 A
Show answer and explanation

Answer: (B) 5 A

For a superposition of a DC component (I_dc) and a sinusoidal AC component of amplitude I₀, the effective current is I_eff = √(I_dc² + I_rms²). Here, I_rms = 3√2 / √2 = 3 A. Therefore, I_eff = √(4² + 3²) = 5 A.

easy

7. The rms value of an alternating voltage is 230 V. Its peak value is:

  1. 325 V
  2. 163 V
  3. 460 V
  4. 147 V
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Answer: (A) 325 V

For a sinusoidal voltage V_rms = V₀/√2, so V₀ = √2 × V_rms = 1.414 × 230 = 325 V.

easy

8. The alternating current flowing through a circuit is given by I = 10√2 sin(50πt - π/4) A. The rms value and frequency of the current are respectively:

  1. 5√2 A and 25 Hz
  2. 10√2 A and 50 Hz
  3. 10 A and 50 Hz
  4. 10 A and 25 Hz
Show answer and explanation

Answer: (D) 10 A and 25 Hz

Comparing with I = I₀ sin(2πft + φ), the peak value I₀ = 10√2 A. Thus, the rms value is I_rms = I₀ / √2 = (10√2) / √2 = 10 A. The angular frequency is ω = 2πf = 50π rad/s, which gives the frequency f = 25 Hz.

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9. Statement I: A capacitor blocks direct current (DC) in the steady state. Statement II: The capacitive reactance is inversely proportional to frequency and tends to infinity as frequency approaches zero.

  1. Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I
  2. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
  3. Statement I is correct but Statement II is incorrect
  4. Statement I is incorrect but Statement II is correct
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Answer: (A) Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Since X_C = 1/(2πfC), as f → 0 for DC, X_C → ∞. Hence, a capacitor offers infinite reactance to direct current in the steady state, completely blocking it.

easy

10. The potential difference across an AC circuit is given by V = 120 sin(120πt) V, where t is in seconds. What is the time taken by the voltage to rise from 0 to +60 V for the first time?

  1. 1/720 s
  2. 1/360 s
  3. 1/240 s
  4. 1/120 s
Show answer and explanation

Answer: (A) 1/720 s

Given V(t) = 120 sin(120πt). For V = 60 V, we have 60 = 120 sin(120πt), which gives sin(120πt) = 1/2. For the first time, 120πt = π/6. Solving for t yields t = (π/6) / (120π) = 1/720 s.

medium

11. A periodic current waveform is defined over one complete cycle of period T as follows: i(t) = +4 A for 0 ≤ t < T/2, and i(t) = -2 A for T/2 ≤ t < T. The root mean square (rms) value of this current is:

  1. 3 A
  2. √10 A
  3. √6 A
  4. 2√2 A
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Answer: (B) √10 A

The mean of the squared current over one cycle is given by <i²> = (1/T) [∫₀^(T/2) (4)² dt + ∫_(T/2)^T (-2)² dt] = (1/T) [16(T/2) + 4(T/2)] = (8 + 2) = 10 A². Taking the square root gives I_rms = √<i²> = √10 A.

easy

12. When the frequency of an AC source connected across an ideal inductor is tripled while keeping the peak voltage constant, the peak current through the inductor becomes:

  1. Three times its initial value
  2. Unchanged
  3. Nine times its initial value
  4. One-third of its initial value
Show answer and explanation

Answer: (D) One-third of its initial value

Inductive reactance is given by X_L = 2πfL. When frequency f is tripled, X_L triples. Since peak current I₀ = V₀ / X_L, tripling the reactance reduces the current to one-third of its original value.

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